1. Ch. 11: Introduction to Compressible Flow
Focus on 1-dimensional, compressible, inviscid flows
Liquids,  = constant for us (1% increase in  for every 1.6 km deep)
Air, 1% change for every 26 m deep
M = 0.3 ~ 5% = /; M = 0.3 ~ 100 m/s or 230 mph
Significant density changes imply significant compression
or expansion work on the gas, which can change T, e, s, …
Compressibility: fluid acceleration because of friction, fluid deceleration in a converging duct, fluid temperature decrease
with heating
Ideal Gas: p = RT (simple, good approximations for our engineering
applications, captures trends)

2. Smits/A Physical Introduction To Fluid Mechanics

3. Density gradients will affect how light is transmitted though
medium (by affecting index of refraction). By applying the
Gladstone-Dale formula it becomes evident that the shadowgraph
is sensitive to changes in the 2nd derivative of the gas density.
Strength of shock can be related to width of dark band.
- Methods of Experimental Physics – Vol 18, Martin
Deflection of light caused by shock compressed gas ahead of a
sphere flying at supersonic speed.

4. Thoughts on the increased complexity of incompressible flow

5. GOVERNING EQUATIONS FOR NEWTONIAN FLUIDS
INCOMPRESSIBLE
/t + /xk(uk) = becomes uk/xk= 0
uj/t + ukuk/xk =
-p/xj+/xj( uk/xk)+/xi[(ui/xj+uj/xi)]+fj
becomes
uj/t + ukuj/xk = -p/xj + (2ui/xjxj) + fj
4 Equations: continuity and three momentum
4 Unknowns: p, u, v, w
Know: , , fj

6. GOVERNING EQUATIONS FOR NEWTONIAN FLUIDS
COMPRESSIBLE
/t + /xk(uk) = 0
uj/t + uk/xk =
-p/xj +/xj( uk/xk) + /xk[(ui/xk + (uj/xi)] + fj
p = p(,T) Thermal ~ p = RT
e = e(,T) Caloric ~ e = CvT
e/t + uke/xk =
-puk/xj +/xj(k T/xj) + (uk/xk)2 + (ui/xk + uj/xi)(uk/xk)
e is the
internal energy
7 Equations: continuity, momentum(3), energy, thermal, state
7 Unknowns: p, u, v, w, e, T, 
Know: , fj, , k

7. Thoughts on the speed of sound

8. COMPRESSIBLE FLOW
front
c2 = (p/)s

9. COMPRESSIBLE FLOW front
If fluid incompressible, gas would behave like solid body and move
everywhere at piston speed. If pressure disturbance is small relative
to p1 then “front” propagates at speed of sound. If large shock waves
occur where speed, temperature, density and pressure change
significantly across shock. (Speed of shock is between the speed of
sound in the compressed and undisturbed gas.)

10. dp/(p/cont) – (const k) k -k-1 d = 0 dp/p – kd/ = 0
~ SPEED OF SOUND ~
Sound waves are pressure disturbances << ambient pressure.
For loud noise: p ~ 1Pa whereas ambient pressure is 105 Pa
Speed of sound: c2 = (p/)s
Assumptions: ideal gas & isentropic
p/k = const, or differentiating
dp/k – pk -k-1 d = 0
dp/(p/cont) – (const k) k -k-1 d = 0
dp/p – kd/ = 0

11. p = RT for ideal gas then c2 = kRT
~ SPEED OF SOUND ~
dp/p – kd/ = 0
dp/d = kp/
p = RT for ideal gas then c2 = kRT
For 20oC and 1 atmosphere
c = 343 m/s = 1126 ft/s = 768 mph

12. M = 1
Dynamic Pressure  Static Pressure

13. M2 = V2/c2 = V2/kRT (ideal gas) p = RT (ideal gas)
M2 = 2(1/2V2/kRT) = 2(1/2 V2/(kp/))
M2 = 2[1/2 V2/(kp)] ~ 1/2 V2/p
M2 ~ dynamic pressure/ static pressure

14. Thoughts on the speed of sound
As related to the speed of the source

15. Regimes of flow:
Acoustics – fluid velocities << c, speed of sound;
fractional changes in p, T and  are important.
(2) Incompressible flow – fluid velocities < c, speed of sound;
fractional changes in  are not significant; fractional changes
in p and T are very important
(3) Compressible flow (gas dynamics) – fluid velocities ~ c,
speed of sound; fractional changes in p, T and  are all
important.
(a) ~ 10 diameters for cylinder; equations are elliptic

16. Propagation of Sound Waves from a Moving Source
SUB SONIC SUPER
(a) ~ 10 diameters for cylinder; equations are elliptic
Propagation of Sound Waves from a Moving Source

17. Some Assumptions

18. It is assumed that the system is always in equilibrium.

19. It has been found by experiment that as
long as the temperatures and pressures
are not too extreme, the flow attains an
instantaneous equilibrium. This continues
to hold even inside shock waves. For all
the flows examined here, all systems will
be assumed to be in equilibrium at all times.
p1,1,
T1,
s1, h1
p2,2,
T2,
s2, h2

20. p =  RT It is assumed that all gases obey ideal gas law:
Kelvin (or Rankine)
Not gauge pressure
R = Ru/Mm = m2/(s2-K) = (N-m)/(kg-K) = J/(kg-K)
R = ft2/(s2-R)

21. Conservation of Energy
(note that u is now the internal energy)

22. Q/m + W/m = E/m; q + w = u
FIRST LAW OF THERMODYNAMICS
Q + W = E = (KE + PE + U)
Q/m + W/m = E/m; q + w = u
W = - pdV
U, internal energy, is energy stored in molecular bonding forces and random molecular motion (KE and PE we will ignore)

23. Ideal gas is composed of point particles which exhibit perfect elastic collisions. Thus internal energy is a function of temperature only. U = f(T)
Enthalpy, h, defined as:
h = u + pv ; h = f(T) since
h(T) = u(T) + RT

24. Specific Heat for Ideal Gas
dQ = mCv,pdT

25. Specific heat is defined as the
amount of heat required to raise
the temperature of a unit mass
of substance by1oK. Different
for constant volume or pressure.

26. Definition of heat capacity at constant volume:
mCvdT = dQ or CvdT = dq
dq + dw = du
if Vol constant, w = -pdv = 0, then dq = du,
Cv = du/dT
“It can be shown that du = Cv dT even
if volume not held constant!!”
- pg 41, Thermal-Fluid Engineering, Warhaft

27. ~ again can be shown to be true even if pressure is not constant!!
Definition of heat capacity at constant pressure:
mCpdT = dQ or Cp = dq/dT
dq + dw = dq – pdv = du
h = u + pv
dh = du +pdv +vdp
if pressure constant, dh = du + pdv = dq
Cp = dh/dT
~ again can be shown to be true even if pressure is not constant!!

28. h = u + pv = u + RT* dh = du + RdT dh/dT = du/dT + R Cp – Cv = R*
Cv = du/dT*
Cp = dh/dT*
h = u + pv = u + RT*
dh = du + RdT
dh/dT = du/dT + R
Cp – Cv = R*
* IDEAL GAS

29. cp/cv = k cp – cv = R cp/cp –cv/cp = R/cp 1 – 1/k = R/cp
cp = R/(1-1/k) = kR/(k-1)

30. cp/cv = k cp – cv = R cp/cv –cv/cv = R/cv k – 1 = R/cv
cv = R/(k – 1)

31. Another Assumption

32. It is assumed that cp/cv
is not a function of T
calorically perfect
For a perfect gas
cp/cv = 1.4

33. cp/cv = k is not a function of temperature

34. cp/cv = k = 1.4 for perfect gas

35. The Second Law

36. Tds = du + pdv = dh –vdp always true!
“The second law of thermodynamics can be stated in several
ways, none of which is easy to understand.”
– Smits, A Physical Introduction to Fluid Mechanics
DEFINITION
S = rev Q/T or dS = (Q/T)rev
dq = du + pdv
Tds = du + pdv = dh –vdp always true!

37. S = rev Q/T or dS = (Q/T)rev
DEFINITION
S = rev Q/T or dS = (Q/T)rev
Change in entropy intimately connected with the
concept of reversibility – for a reversible, adiabatic
process entropy remains constant.
For any other process the entropy increases.

38. + ideal gas and constant specific heats
What we can do with:
Tds = du + pdv = dh –vdp
+ ideal gas and constant specific heats

39. Tds = du + pdv = dh –vdp ds = du/T + RTdv/T
Cv = du/dT Ideal Gas Cp = dh/dT
p = RT = (1/v)RT
Tds = du + pdv = dh –vdp
ds = du/T + RTdv/T
ds = CvdT/T + (R/v)dv
s2 – s1 = Cvln(T2/T1) + Rln(v2/v1)
s2 – s1 = Cvln(T2/T1) - Rln(2/1)

40. Cp – Cv = R; R/Cv = k – 1 s2 – s1 = Cvln(T2/T1) - Rln(2/1)
If isentropic s2 – s1 = 0
ln(T2/T1)Cv = ln(2/1)R
Cp – Cv = R; R/Cv = k – 1
2/1 = (T2/T1)Cv/R = (T2/T1)1/(k-1)
assumptions
ISENROPIC & IDEAL GAS
& constant cp, cv

41. Tds = du + pdv = dh –vdp ds = dh/T - vdp
Cv = du/dT Ideal Gas Cp = dh/dT
p = RT = (1/v)RT
Tds = du + pdv = dh –vdp
ds = dh/T - vdp
ds = CpdT/T - (RT/[pT])dp
s2 – s1 = Cpln(T2/T1) - Rln(p2/p1)
s2 – s1 = Cpln(T2/T1) - Rln(p2/p1)

42. Cp – Cv = R; R/Cp = 1- 1/k s2 – s1 = Cpln(T2/T1) - Rln(p2/p1)
If isentropic s2 – s1 = 0
ln(T2/T1)Cp = ln(p2/p1)R
Cp – Cv = R; R/Cp = 1- 1/k
p2/p1 = (T2/T1)Cp/R = (T2/T1)k/(k-1)
assumptions
ISENROPIC & IDEAL GAS
& constant cp, cv

43. Stagnation Reference (V=0)

44. BE: 1-D, energy equation for adiabatic and no shaft or viscous work.
(p2/2) + u2 + ½ V22 + gz2 = (p1/1) + u1 + ½ V12 + gz1
Definition: h = u + pv = u + p/; assume z2 = z1
h2 + ½ V22 = h1 + ½ V12 = ho + 0
Cp = dh/dT (ideal gas)
ho – h1 = cp (To – T) = ½ V12
T0 = ½ V12/cp + T = T (1 + V2/[2cpT])

45. T0 = T (1 + V2/[2(kR/(k-1)]T])
T0 = ½ V12/cp + T = T (1 + V2/[2cpT])
cp = kR/(k-1)
T0 = T (1 + V2/[2(kR/(k-1)]T])
T0 = T (1 + (k-1)V2/[2kRT])
c2 = kRT
T0 = T (1 + (k-1)V2/[2c2])
V2/ c2
T0 = T (1 + [(k-1)/2] M2)

46. To/T = 1 + {(k-1)/2} M2 STEADY, 1-D, ENERGY EQUATION FOR
ADIABATIC FLOW OF A PERFECT GAS

47. Ideal gas and isentropic
/o = (T/To)1/(k-1)
To/T = 1 + {(k-1)/2} M2
/o = (1 + {(k-1)/2} M2 )1/(k-1)
Ideal gas and isentropic
and constant cp, cv
(isentropic = adiabatic + reversible)

48. (isentropic = adiabatic + reversible)
p/p0 = (T/To)k/(k-1)
To/T = 1 + {(k-1)/2} M2
p/p0 = (1 + {(k-1)/2} M2)k/(k-1)
Ideal gas and isentropic
and constant cp, cv
(isentropic = adiabatic + reversible)

49. QUIZ When a fixed mass of air is heated from 20oC to 100oC –
What is the change in enthalpy?
For a constant volume process,
what is the change in entropy?
For a constant pressure process,
For an isentropic process
what are the changes in p and ?
Compare speed of sound
for isentropic and isothermal conditions.

50. h2 – h1 = Cp(T2- T1)
s2 – s1 = Cvln(T2/T1)
s2 – s1 = Cpln(T2/T1)
100/ 20 = (T100/T20)2.5
2.5 = 1/(k-1) k = 1.4 for ideal gas
p100 / p20 = (T100/T20)3.5
3.5 = k/(k-1) k = 1.4 for ideal gas

51. (e) c2 = {dp/d}
But c2 = (p/)|T does not equal c2 = (p/|S)
If isentropic p/k = constant (ideal gas)
Then c = {(p/)|S}1/2 = (kRT)1/2
= (1.4 * * ( ))1/2
= m/s
If isothermal p = RT (ideal gas)
Then c = {(p/)|T}1/2 = (RT)1/2
= ( X ( )1/2
= m/s 18% too low